MYSQL：仅根据group by ID在第一行添加一些数据

我有这组样本数据：

表格发票：

表1\u插件：

表顺序：

我想要达到的目标：

----------------------------------------------------------------------------------------------------------------------------------
| trans_date  | inv_number |  box_name   |   size   |  gross   |  box_number  | others | description                     |  net  |
----------------------------------------------------------------------------------------------------------------------------------
| 2017-03-24  | 0000000173 | Jumbo Box   | 23x25x17 |  112.00  |  FCI107056   | -24.00 |  Price Adjust. - Jumbo -12.00   | 88.00 |
|             |            |             |          |          |              |        |  Price Adjust. - Regular -12.00 |       |
----------------------------------------------------------------------------------------------------------------------------------
| 2017-03-24  | 0000000173 | Regular Box | 20x23x17 |  102.00  |  FCI107057   |      0 |  NULL                           | 102.00|
----------------------------------------------------------------------------------------------------------------------------------

我当前的查询：

SELECT DATE(i.date) AS trans_date, i.invoice_number AS inv_number, o.box_name as box_name, 
o.size AS size, (SELECT Price + Tax FROM orders WHERE ID = o.ID) AS gross, 
o.box_number AS box_number,
SUM(a.amount) AS others, 
(SELECT GROUP_CONCAT(CONCAT(description, ' ', amount) SEPARATOR '<br />') FROM invoice_addon WHERE invoice_id = i.ID) AS description, 
(SUM(o.price + o.tax) + SUM(a.amount)) AS net 
FROM `invoice` i 
INNER JOIN orders o ON i.ID = o.invoice_id 
LEFT JOIN invoice_addon a ON i.ID = a.invoice_id 
WHERE i.ID = 81
GROUP BY o.ID

可能吗？我怎样才能做到这一点？ 我在MySQL中使用CodeIgniter和DataTable来实现这一点，您需要一些可以存储当前行号并检查其是否等于1的变量。大概是这样的：

SET @row_number = 0;
SELECT DATE(i.date) AS trans_date, i.invoice_number AS inv_number, o.box_name as box_name, 
o.size AS size, (SELECT Price + Tax FROM orders WHERE ID = o.ID) AS gross, 
o.box_number AS box_number,
if((@row_number:=@row_number + 1)=1,SUM(a.amount), null) AS others, 
if(@row_number=1,(SELECT GROUP_CONCAT(CONCAT(description, ' ', amount) 
SEPARATOR '<br />') FROM invoice_addon WHERE invoice_id = i.ID), null) AS description, 
(SUM(o.price + o.tax) + SUM(a.amount)) AS net 
FROM `invoice` i 
INNER JOIN orders o ON i.ID = o.invoice_id 
LEFT JOIN invoice_addon a ON i.ID = a.invoice_id  
WHERE i.ID = 81
GROUP BY o.ID

---

主要的问题是没有invoice_插件的链接，因此例程知道它与哪个框相关。你应该重新找回你的梦想approach@nbk只要他们只在第一个数据上列出，这并不重要。总是发票id对于orders表和invoice\u addon都是通用的吗？无法做到这一点？因为您没有为invoice\u addon.id定义关系，所以没有算法知道应该在何处添加发票税，所以建立ar关系，一切都很容易。谢谢您的回答。它适用于特定的数据i.ID=81，但我希望使用它时不带where子句，因此在每次出现新发票时，都应填写其他和说明。\u ID.@Elimariaa从您的question@elimariaaa将最后一张发票的id存储在某个var中，并检查是否等于新发票的id是，对不起。谢谢

----------------------------------------------------------------------------------------------------------------------------------
| trans_date  | inv_number |  box_name   |   size   |  gross   |  box_number  | others | description                     |  net  |
----------------------------------------------------------------------------------------------------------------------------------
| 2017-03-24  | 0000000173 | Jumbo Box   | 23x25x17 |  112.00  |  FCI107056   | -24.00 |  Price Adjust. - Jumbo -12.00   | 88.00 |
|             |            |             |          |          |              |        |  Price Adjust. - Regular -12.00 |       |
----------------------------------------------------------------------------------------------------------------------------------
| 2017-03-24  | 0000000173 | Regular Box | 20x23x17 |  102.00  |  FCI107057   |      0 |  NULL                           | 102.00|
----------------------------------------------------------------------------------------------------------------------------------

SELECT DATE(i.date) AS trans_date, i.invoice_number AS inv_number, o.box_name as box_name, 
o.size AS size, (SELECT Price + Tax FROM orders WHERE ID = o.ID) AS gross, 
o.box_number AS box_number,
SUM(a.amount) AS others, 
(SELECT GROUP_CONCAT(CONCAT(description, ' ', amount) SEPARATOR '<br />') FROM invoice_addon WHERE invoice_id = i.ID) AS description, 
(SUM(o.price + o.tax) + SUM(a.amount)) AS net 
FROM `invoice` i 
INNER JOIN orders o ON i.ID = o.invoice_id 
LEFT JOIN invoice_addon a ON i.ID = a.invoice_id 
WHERE i.ID = 81
GROUP BY o.ID

----------------------------------------------------------------------------------------------------------------------------------
| trans_date  | inv_number |  box_name   |   size   |  gross   |  box_number  | others | description                     |  net   |
----------------------------------------------------------------------------------------------------------------------------------
| 2017-03-24  | 0000000173 | Jumbo Box   | 23x25x17 |  112.00  |  FCI107056   | -24.00 |  Price Adjust. - Jumbo -12.00   | 200.00 |
|             |            |             |          |          |              |        |  Price Adjust. - Regular -12.00 |        |
----------------------------------------------------------------------------------------------------------------------------------
| 2017-03-24  | 0000000173 | Regular Box | 20x23x17 |  102.00  |  FCI107057   | -24.00 |  Price Adjust. - Jumbo -12.00   | 180.00 |
|             |            |             |          |          |              |        |  Price Adjust. - Regular -12.00 |        |
----------------------------------------------------------------------------------------------------------------------------------

SET @row_number = 0;
SELECT DATE(i.date) AS trans_date, i.invoice_number AS inv_number, o.box_name as box_name, 
o.size AS size, (SELECT Price + Tax FROM orders WHERE ID = o.ID) AS gross, 
o.box_number AS box_number,
if((@row_number:=@row_number + 1)=1,SUM(a.amount), null) AS others, 
if(@row_number=1,(SELECT GROUP_CONCAT(CONCAT(description, ' ', amount) 
SEPARATOR '<br />') FROM invoice_addon WHERE invoice_id = i.ID), null) AS description, 
(SUM(o.price + o.tax) + SUM(a.amount)) AS net 
FROM `invoice` i 
INNER JOIN orders o ON i.ID = o.invoice_id 
LEFT JOIN invoice_addon a ON i.ID = a.invoice_id  
WHERE i.ID = 81
GROUP BY o.ID